Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000!

Question

Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000!
🧐 Not the exact question you are looking for?Go ask a question

Solution 1

To prove this, we will use the concept of mathematical induction.

Step 1: Base Case We are given that 20236000 ≤ 6000!. Since 2023 < 20236000, we can say that 2023*6000 ≤ 6000!. So, we can take our base case as N = 6000.

Step 2: Inductive Step Assume that the statement is true for some k ≥ N, i.e. Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study prob

Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solv

This problem has been solved

Similar Questions

Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000!

Prove that there exists a positive integer N such that for all n ≥ N , we have2023n ≤ n!. You may use, without proof, the fact that 20236000 ≤ 6000

Prove that there exists a positive integer N such that for all n ≥ N, we have 2023^n ≤ n!.

rational number for all distinct numbers a, b. e in A. Prove that there exists a positive integer M such that a √M is a

Prove that for every positive integer n, and sets A and B1, B2, ..., Bn, A - ∩(i=1 to n) Bi = ∪(i=1 to n) (A - Bi).

1/3